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At the corner of the cube circular arcs are drawn and the area enclosed shaded. What fraction of the surface area of the cube is shaded? Try working out the answer without recourse to pencil and paper.

According to Plutarch, the Greeks found all the rectangles with
integer sides, whose areas are equal to their perimeters. Can you
find them? What rectangular boxes, with integer sides, have their
surface areas equal to their volumes?

Is it possible to remove ten unit cubes from a 3 by 3 by 3 cube made from 27 unit cubes so that the surface area of the remaining solid is the same as the surface area of the original 3 by 3 by 3 cube?

Cuboids

Stage: 3 Challenge Level:

Steven and Ryan found that a 2x4x7
cuboid has a surface area of 100 square units.

Justin from Mason Middle School found
that a 1x2x16 cuboid also satisfies the conditions:

I back it up by saying that
$1\times 2 = 2$,
$2 \times16 = 32$
and $16 \times1 = 16$
and add them all up and you get $50$, and then multiply by $2$ and
you get a square area of $100$!

Megan was able to show that these were the
only two possible solutions:

Call the lengths of the 3 dimensions (height, depth, width) $x$,
$y$ and $z$.
The surface area is $2xy+2xz+2yz$, as the area of each face is
calculated by multiplying its two sides together, and there are $2$
of each face.
Hence, $2xy+2xz+2yz = 100$.
Dividing by $2$, $xy+yz+xz = 50$.
I will assume that $x$ is the shortest side and $z$ the longest to
avoid repeating solutions. Therefore I must find integer solutions
to the equation $xy+yz+xz = 50$ where $x < y < z$.
Rearranging the equation,
$yz+xz = 50-xy$
$z(x+y) = 50-xy$
$z = (50-xy)/(x+y)$.
I used an excel spreadsheet with 3 columns, 1 for each of $x$, $y$
and $z$.
In the $z$ column I write the rearranged formula.
I then started from $x = 1$, $y = 1,2,3...$ looking for integer
values of $z$ until I reached a solution which had been repeated
(as here $y$ is bigger than $z$, so I would be repeating values
with $y$ and $z$ swapped around) or where $z$ became smaller than
$y$.
For $x = 1$ I found only 1 solution, $(1, 2, 16)$.
Checking with the original formula this does agree to a total of
$100$.
I then continued repeating the procedure with $x = 2$, $y =
2$,$3$,$4$... and found a solution of $(2, 4, 7)$.
With $x = 3$, $y = 3$,$4$,$5$... there are no solutions, as $z$
becomes smaller than $y$ where $y=5$ (and $z = 4.375$) and there
are no integer solutions before this.
With $x = 4$, $y = 4,5,6...$ $z$ becomes smaller than $y$ when $y =
5$ $(z = 3.333)$ therefore there are no solutions where $x = 4$.
After $x = 4$, the intitial value of $z$ is always smaller than
$y$, therefore there are no further solutions.
The only 2 solutions to the problem are: $(1, 2, 16)$ and $(2, 4,
7)$.

Fred from Albion Heights School also offered
some non-integer solutions:

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